Trigonometry: Find $\sin \theta$ when $\tan \theta$ is known.

Question

if $\tan \theta = \sqrt{63}$ and $\cos \theta$ is negative, find $\sin \theta$.

So since $\tan \theta$ is positive and $\cos \theta$ is negative, it lies in the $3$^rd quadrant. So $\sin$ is negative, but I don't know how to find $\sin \theta$, please guide me... Thank you.

Answer 1

HINT

Remember $\tan \theta$ is just $\frac{\sin \theta}{\cos \theta}$. Here you have $\tan \theta = \frac{\sqrt{63}}{1}$ Now, draw a triangle with the sides as $\sqrt{63}$ and $1$. Now you should be able to find $\sin \theta$ and adjust the signs.

FURTHER HINT

Answer 2

In general $$\tan^2A=\frac{\sin^2A}{\cos^2A}=\sin^2A\cdot\sec^2A\iff\sin^2A=\frac{\tan^2A}{\sec^2A}=\frac{\tan^2A}{1+\tan^2A}$$

Answer 3

$\tan \theta = \sqrt{63} =\frac {\sqrt{63}}{1} = \frac {y-ordinate}{x-ordinate} =\frac {-\sqrt{63}}{-1}$; since it lies in the 3rd quadrant.

Then, $\sin \theta = \frac {y-ordinate}{radius} = \frac {-\sqrt{63}}{8}$

$\sqrt (x-ordinate^2 + y-ordinate^2)$ = 8 = radius, which is always positive.

Answer 4

$\text{Because, }\tan{\theta}=\tan{(\pi+\theta)}$ $$\tan{(\pi+\theta)}=\sqrt{63}$$ $\text{Because, }\sin{(\arctan{x})}=\frac{x\sqrt{1+x^2}}{1+x^2}$ $$\sin{(\pi+\theta)}=\frac{3\sqrt{7}}{8}$$ $$\sin{\theta}=-\frac{3\sqrt{7}}{8}$$ $\text{Note that, }\cos{(\arcsin{(-\frac{3\sqrt{7}}{8})})}\text{ could be smaller than 0. }$